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I am not very experienced in mathematical notation, so please excuse some terminology misuse or formatting shortcomings.

I have a project in which a value needs to increase from a set minimum to a set maximum in a set number of seconds. It is easy to calculate the linear value based on ratios.

Let $v$ = desired value, $n$ = minimum limit, $x$ = maximum limit, $t$ = elapsed time, and $t_x$ = allocated time:

$$v = \frac{t}{t_x}(x-n) + n.$$

Thus if my values are:

$$n = 5, x = 90, t_x = 1800 \text{ (half hour)}$$

For elapsed time of $5$ minutes ($600$ s):

$$v = \frac{600}{1800} (90-5) + 5 = 33.3.$$

The problem is I want to change this linear growth to exponential growth, and I'm not sure how to alter the formula.

So instead of $33.3$ at $5$ minutes, I would rather have $13$ for example. (Slow initial change, rapid later change.)

How can I insert an exponential growth factor into my equation and honor the minimum and maximum values allowed?

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3 Answers 3

up vote 3 down vote accepted

I will change notation slightly. Our initial smallest value is $a$, and our largest value, after say $k$ seconds, is $b$. So every second our amount gets multiplied by $(b/a)^{1/k}$, the $k$-th root of $b/a$. At elapsed time $t$ seconds, where $0 \le t \le k$, the value is $$a \left(\frac{b}{a}\right)^\frac{t}{k}.$$

This is what would happen if we have an initial population $a$ of bacteria, growing under ideal conditions, and ending up with population $b$ after $k$ seconds. The formula above gives the population at time $t$, where $0 \le t \le k$.

It is also what happens if we have an initial amount of money $a$, which under continuous compounding grows to $b$ in time $k$.

Remark: The quantity $Q$ grows exponentially if and only if the quantity $\log Q$ grows linearly. So alternately, you could translate your knowledge about linear growth to a formula about exponential growth.

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Your explanation was extremely helpful as well as a clear presentation of the formula. I admit I am more encouraged to return to studying mathematics! –  JYelton Aug 1 '12 at 17:28

Let your model be $v(t) = v_0 e^{\alpha t}$, where $v_0$ and $\alpha$ are constants to be determined.

From your data, you want $v(0) = n$, and $v(t_x) = x$.

This immediately gives $v_0 = n$, and then we have $v(t_x) = n e^{\alpha t_x}$, from which we get $\alpha = \frac{1}{t_x} \ln \frac{v(t_x)}{n} = \frac{1}{t_x} \ln \frac{x}{n}$.

So the model is $$ v(t) = n e ^ { \frac{t}{t_x} \ln \frac {x}{n}}.$$

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I see several sorts of ways to proceed. So let's suppose we had minimum $5$, max $90$, and $1800$ seconds to get there. Then we might have an exponential of the form $Pe^{rt}$ or perhaps $Pe^{rt} + c$. We might choose $f(t) = e^{rt} + 4$, so that $f(0) = 5$ (the minimum). We ask ourselves what $r$ would make it so that $f(1800) = 90$, the max?

Then we'd look at $e^{r1800} = 90 - 4 = 86$, or $\ln 86/1800 = r$ (that's really small). This would give the start of the exponential, growing slowly and then moreso.

But every choice was arbitrary, so let's look at a few other things you might do. Perhaps you want to change the rate of growth. You might do $5e^{rt}$, finding the correct $r$ again. Or you might do something like $Pe^{r(t - 10)}$, shifting along the exponential curve to places that change more slowly, etc.

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